In ΔHIJ, the measure of ∠j=33cm,∠H=148°and I= 21°Find the length of i to the nearest tenth of a centimeter

i = 62 cm

Step-by-step explanation:

I have drawn the triangle to reflect the angles and dimension given and attached it.

The missing angle in the triangle is ∠J.

we know that sum of angles in a triangle = 180°

Thus;

148 + 21 + ∠J = 180

∠J = 180 - 148 - 21

∠J = 11°

To find i, we will make use of sine rule which is; a/sin A = b/sin B = c/sin C

Thus;

i/sin I = j/sin J

i = j sin I/sin J

i = (33 × sin 21)/sin 11

i = 61. 979 cm

To the nearest tenth gives;

i = 62 cm

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!